Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
Two different galvanometers G1 and G2, have internal resistances r1and r2. The galvanometers G1 and G2 require the same current IC1=IC2 for a full-scale deflection of their pointers. These galvanometers G1 and G2 are used to build lab-made ammeters A1 and A2 . Both ammeters A1 and A2 have the same maximum scale reading Imax1=Imax2=Imax. To build A1 ,shunt resistor of resistance Rsh1is used and to build A2 , shunt resistor of resistance Rsh2 is used. The value of these shunt resistor resistances are such that: Rsh1=3Rsh2. What is the ratio oftheir internal resistances: r1:r2?
Answer:
there are 3 photos attached. so check
Explanation:
01.04 Law of Conservation of Energy
science question
Answer:
law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
Cho hai mặt cầu đồng tâm O tích điện đều. Bán kính của hai mặt cầu lần lượt là R1 và R2 (R2>R1). Điện tích mặt trong là q và mặt ngoài là Q
Tính cường độ điện trường tại một điểm cách tâm O một đoạn r (biết R1 < r < R2)
Tính hiệu điện thế giữa hai mặt cầu
Answer:
you will stc ohxoyxct txxtx xigigjjgjvvixiffjz,iffzikzfjvixii. hi h ohigiogooigoh
Explanation:
k jjvhvojvovvojivivivihvi hj
A student of mass 50kg takes 15seconds to run up a flight of 50 steps. If each step is 20cm, calculate the potential energy of the student at the maximum height
Answer:
the answer is 49000 joules at the maximum height
Explanation:
we know the mass (50kg)
we know the acceleration due to gravity(9.8m/s²)
we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m
the formula is potential energy=mgh
m for mass
g for acceleration due to gravity
h for height
multiply them
50x9.8x100
we get 49000
the unit of potential energy is joules so the answer is
49000 joules
Answer:
49000 joules
Explanation:
hope it helpss
Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
explain why our sweat is salty?
Answer:
Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.
Hope this helps..
A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?
Answer:
0.032 [tex]m/s^2[/tex]
Explanation:
Given :
Weight of the astronaut = 69 kg
Weight of the physics book = 28 kg
Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]
The force that is applied on the astronaut :
[tex]F=ma[/tex]
[tex]$=69 \times 0.013$[/tex]
= 0.897 N
Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N
Therefore, the acceleration of the physics book is given by :
[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]
[tex]$a = \frac{0.897}{28}$[/tex]
a = 0.032 [tex]m/s^2[/tex]
Hence, the acceleration of the physics book is 0.032 [tex]m/s^2[/tex].
Answer:
The acceleration of astronaut is 5.27 x 10^-3 m/s^2.
Explanation:
mass of astronaut, M = 69 kg
Mass of book, m = 28 kg
acceleration of book, a = 0.013 m/s^2
Let the acceleration of astronaut is A.
According to the Newton's third law, for every action there is an equal and opposite reaction.
So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.
M A = m a
69 x A = 28 x 0.013
A = 5.27 x 10^-3 m/s^2
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
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When you have a straight horizontal line on a velocity time graph, what does this tell you about the object’s motion in terms of velocity and acceleration?
Answer:
It tell you that the velocity is constant, what means that there's no acceleration
A rabbit is moving in the positive x-direction at 2.70 m/s when it spots a predator and accelerates to a velocity of 13.3 m/s along the positive y-axis, all in 2.40 s. Determine the x-component and the y-component of the rabbit's acceleration.
Answer:
the answer is nearly 5.655 [tex]ms^{-2}[/tex]
Explanation:
Given,
[tex]v_{x}=2.7 ms^{-1}[/tex]
[tex]v_{y}=13.3 ms^{-1}[/tex]
[tex]t=2.4 s[/tex]
[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as [tex]a=\frac{v-u}{t}[/tex])
[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]
[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]
[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]
[tex]=5.655 ms^{-2}[/tex]
hope you have understood this...
pls mark my answer as the brainliest
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:
Answer:
Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)
Explanation:
Time=Distance/Speed
=14.5/6.63
=2.19s(to 3s.f.)
Acceleration=Final Velocity(v)-Initial Velocity(u)/Time
9.81=v-6.63/2.19
v-6.63=21.5
v=28.1m/s
how much heat is produced in one hour by an electric iron which draws 2.5ampere when connected to a 100V supply
Explanation:
I=2.5 Ampere ; V=100V ;t = 1 hour=60secs
We know Heat = VIt
H=100×2.5×60=15,000J
You are trying to push a 14 kg canoe across a beach to get it to a lake. Initially, the canoe is at rest, and you exert a force over a distance of 5 m until it has a speed of 2.1 m/s.
a. How much work was done on the canoe?
b. The coefficient of kinetic friction between the canoe and the beach is 0.2. How much work was done by friction on the canoe?
c. How much work did you perform on the canoe?
d. What force did you apply to the canoe?
Answer:
I hope this is correct!
Explanation:
a) work = (force)(displacement)
We know that d = 5m, now we just need to find the force
We need to calculate the acceleration first using a = v^2 - u^2 / 2d
final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m
a = 2.1^2 - 0^2 / 2(5)
= 0.441 m/s^2
Now we can find force using f = ma
f = (14kg)(0.441m/s^2)
= 6.174 newtons
Finally, you can calculate work now that you have force!
work = (6.174n)(5m)
= 30.87 J (you may need to round sig figs!)
b) Work done by friction = (coefficient of kinetic friction)(mg)(v)
m = 14kg, g = 9.8, coefficient of friction = 0.2
force due to friction = (0.2)(14kg x 9.8)(2.1)
= 57.624 J (again you may need to round to sig figs)
c) Work you perform = total work in direction of motion + work done by friction
total work = 30.87 J, work done by friction = 57.624 J
Work you perform = 30.84 + 57.624
= 88.464 J
d) Force you applied = work you performed / distance
work you performed = 88.464 / 5 m
= 17.6929 N (again you may need to round to sig figs)
Hope this helps ya! Best of luck <3
Which of the following measures is equal to 700 km?
Answer:
1km=1000m
700km=
700×1000=700000
=700000metres
hope this helps
krichoffs law of current questions
Answer:
Explanation:
Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.
#I AM ILLITERATE
A 1.50 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s , and at point B it has slowed to 1.25 m/s .
Part A
How much work was done on the book between A and B ?
Part B
If -0.750J of work is done on the book from B to C , how fast is it moving at point C ?
Part C
How fast would it be moving at C if 0.750J of work were done on it from B to C ?
I assume friction is the only force acting on the book as it slides.
(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²
W ≈ -6.56 J
(B) Using the work-energy theorem again, the speed v of the book at point C is such that
-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v = 0.750 m/s
(C) Take the left side to be positive, then solve again for v.
0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v ≈ 1.60 m/s
The car has a mass of 0·50 kg. The boy
now increases the speed of the car to 6·0
ms-1 . The total radial friction between
the car and the track has a maximum
value of 7.0 N. Show by calculation that
the car cannot continue to travel in the circular path.
Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
What is meant by centripetal force ?Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.
Here,
Mass of the car, m = 0.5 kg
Velocity of the car, v = 6 m/s
Radial friction between the car and the track, f = 7 N
The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.
So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.
So,
mv²/r > f
0.5 x 6²/r > 7
Therefore, the radius of the circular track,
r < 18/7
r < 2.57 m
Hence,
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
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You want to swim from one side of a river to another side. Assume your speed is three miles per hour in the west direction, with negligible water velocity. When you reach a certain point, you will encounter water flow with a velocity of 6.2 miles per hour in the north direction. What is your resultant speed and direction
Answer:
speed = 6.71 mph and angle is 71.2 degree.
Explanation:
speed of person, u = 3 miles per hour
speed of water, v = 6 miles per hour
Resultant speed
[tex]V =\sqrt{v^2 + u^2}\\\\V = \sqrt{3^2 + 6^2}\\\\V = 6.71 mph[/tex]
The angle from the west is
tan A = 6/2 = 3
A = 71.6 degree
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
(c) It takes you hours to to bring the turkey from to . During that time, the electrical grid transfers a constant Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings
Complete Question
(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?
Answer:
[tex]Q=4.50 *10^7J[/tex]
Explanation:
From the question we are told that:
Time [tex]t=5hours[/tex]
Temperature rise [tex]dT= 65\textdegree[/tex]
Power [tex]P=2500.0 Watts[/tex]
Generally, the equation for Power is mathematically given by
[tex]P=\frac{Q}{t}[/tex]
Therefore
[tex]Q=2500*5*360[/tex]
[tex]Q=4.50 *10^7J[/tex]
Which of the following describes an electric conductor
The following describes an electric conductor : A material that has low resistance and allows the charges to move freely. The correct option is D.
What is conductor?A conductor is a material or metal which allows the electrons to flow through it. In other words, a conductor allows the current to pass through them.
A battery also called as the voltage source, provides sufficient voltage or energy to excite electrons in the conductor.
Opposition offered to the flow of current is called as the resistance. The electrical element used in the circuit is the resistor.
So, an electric conductor is a material that has low resistance and allows the charges to move freely.
Thus, the correct option is D.
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If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]
Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
In order to test an intentionally weak adhesive, the bottom of the small 0.15-lb block is coated with adhesive and then the block is pressed onto the turntable with a known force. The turntable starts from rest at t = 0 and uniformly accelerates with a = 2 rad/s^2. If the adhesive fails at exactly t = 3 s, then determine:
a. the magnitude of the ultimate shear force that the adhesive supports
b. the angular displacement of the turntable at the time of failure
Answer:
answer
Explanation:
it is the answer which was presented in the year
A 150g copper bowl contains 220g of water, both at 20.0oC, A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100oC, Neglect energy transfers with the environment.
a) How much energy (in calories) is transfered to the water as heat?
b) How much to the bowl?
c) What is the original temperature of the cylinder?
We have that the energy (in calories) is transferred to the water as heat,to the bowl and the original temperature of the cylinder is mathematically given as
Qw=20.3 kcal Q= 1.11 kcal Ti=873°CEnergy
Generally the equation for the is mathematically given as
(a)
The heat transferred to the H20
Qw= CwMwdT+Lvms
Qw=((220g)(100°C-20.0T)+(539 caVg)(5.00 g)
Qw=20.3 kcal .
(b)
The heat transferred to the bowl is
Qb= CbmbdT
Q= (0.0923 cal/gC)(150g)(100°C-20.0°C)
Q= 1.11 kcal
(c)
original temperature of the cylinder
-Qw- Qb = CcMc(T2-T1)
[tex]T1=\frac{Qw+Qm}{CcMc}+T2[/tex]
T1=873C
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which watch is more preferable for the measurement of time among pendulum, quartz and atomic watch
Answer:
pendulum, quartz
Explanation:
potential diffetence
Answer:
6v
Explanation:
V=IR
V= 2* 3
V= 6 volts
A race car goes from a complete stop at the start line to 150 miles per hour in 5 seconds. What is its acceleration? Show your work.
Answer:
Explanation:
150/5 = 30
30mph per 1 second
RATIO of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
1) 7/29
2) 9/31
3) 5/27
4) 5/23
Answer:
[tex]5/27[/tex]
Explanation:
wavelengths for Lyman series
[tex]\lambda=\frac{1}{R(1-\frac{1}{4} })=\frac{4}{3R}[/tex]
wavelengths for Balmer series
[tex]\lambda_B=\frac{1}{R(\frac{1}{4}-\frac{1}{9}) } =\frac{1}{R(\frac{5}{36}) } =\frac{36}{5R}[/tex]
[tex]\frac{ \lambda_L}{ \lambda_B} =\frac{4}{3R} \times\frac{5R}{36} =5/27[/tex]
OAmalOHopeO
The ratio of longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is 5/27. The correct option is 3.
What is Lyman and Balmer series?
Lyman and Balmer series are sets of spectral lines in the emission spectrum of hydrogen, which result from the transitions of the electron from higher energy levels to lower energy levels.
The Lyman series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=1 energy level. These transitions release energy in the form of ultraviolet photons. The lowest energy level in hydrogen is the n=1 energy level, which is also called the ground state. Therefore, the Lyman series includes the transition of the electron from any energy level greater than or equal to n=2 to the ground state.
The Balmer series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=2 energy level. These transitions release energy in the form of visible photons. The lowest energy level in the Balmer series is the n=2 energy level. Therefore, the Balmer series includes the transition of the electron from any energy level greater than or equal to n=3 to the n=2 energy level.
Lyman and Balmer's series are named after the scientists who discovered them. The Lyman series is named after Theodore Lyman, an American physicist who discovered the series in 1906. The Balmer series is named after Johann Balmer, a Swiss mathematician who discovered the series in 1885.
Here in the Question,
The longest wavelength in the Lyman series of the hydrogen spectrum corresponds to the transition from the n = 2 energy level to the n = 1 energy level, while the longest wavelength in the Balmer series corresponds to the transition from the n = 3 energy level to the n = 2 energy level.
The wavelengths of these transitions can be calculated using the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength of the photon emitted, R is the Rydberg constant (1.097 × 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron.
For the longest wavelength in the Lyman series, we have n1 = 2 and n2 = 1, so:
1/λ_lyman = R(1/2^2 - 1/1^2) = 3R/4
For the longest wavelength in the Balmer series, we have n1 = 3 and n2 = 2, so:
1/λ_balmer = R(1/3^2 - 1/2^2) = 5R/36
Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is:
λ_lyman/λ_balmer = (3R/4)/(5R/36) = 27/20
Simplifying this ratio gives:
λ_lyman/λ_balmer = 27/20
Multiplying both the numerator and denominator by 1/3R, we get:
λ_lyman/λ_balmer = (1/2)/(1/3) = 3/2
Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is 3:2, or 3/5 in fractional form. Simplifying this ratio gives:
λ_lyman/λ_balmer = 5/3
Taking the reciprocal of both sides, we get:
λ_balmer/λ_lyman = 3/5
Therefore, the correct answer is (3) 5/27.
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Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound
The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]
Therefore,
[tex]$AD-BD = \frac{\lambda}{2}[/tex]
[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2 \times 0.4985$[/tex]
[tex]$\lambda = 0.99714 \ m$[/tex]
Thus the frequency is :
[tex]$f=\frac{v}{\lambda}$[/tex]
[tex]$f=\frac{340}{0.99714}$[/tex]
[tex]f=340.9744[/tex] Hz
